Applications or physical interpretation of auto-convolution?

Question

I wonder if anyone has any experience with auto-convolution. In particular i'm interested in understanding the physical interpretation of it. I understand what convolution, correlation and auto-correlation are, also i'm aware that the definition of auto-convolution will be something like $$f\ast f = \int_{-\infty}^{\infty} f(\tau)f(t-\tau)d\tau $$ but i still don't get what are the implications o the meaning of it. I've been looking for a while and so far i haven't found any good or detailed explanation (on constrast with auto-correlation). So, if anyone has any experience dealing with this topic or has any intuitive interpretation that could share, i'll appreciate it. Thanks.

Answer 1

Autoconvolution is used in signal detection, but the way you've written it is not correct. Suppose you're trying to detect a signal $f(t)$ by filtering with h(t).

$y(t) = f(t) \ast h(t)$

You want to maximize your response to the signal $f(t)$. We can do this by maximizing the correlation coefficient between $f$ and $h$. Here the correlation is time-varying , so we'll maximize the average autocorrelation coefficient. We'll assume the $f$ and $h$ signals have DC values of zero for simplicity.I'll use $\mu_y$ to denote average value of a the auto-correlation of response, $y$.

$ h = argmax_{h} \ \ \mu_y\big(E_f E_h \big)^{-\frac{1}{2}}$

Let's take a differential of our performance $ J \propto \mu_y\big(E_f E_h \big)^{-\frac{1}{2}}$ with respect to h

$\partial_{h(\tau)} J \propto \partial_{h(\tau)} \bigg((\mu_y\big(E_f E_h \big)^{-\frac{1}{2}}\bigg)$

$\ \ \ \ \ \ \ \ \ = \frac{1}{|dom(f)|}\bigg(\partial_{h(\tau)}\mu_y\bigg)\big(E_fE_h)^{-\frac{1}{2}} +\frac{1}{2}\mu_yE_f^{-\frac{1}{2}}E_h^{-\frac{3}{2}}\bigg(\partial_{h(\tau)}E_h\bigg)$

$\ \ \ \ \ \ \ \ \ \propto \bigg(\partial_{h(\tau)}\frac{1}{|dom(f)|}\int_{\infty}^{\infty}f(t-\tau)h(\tau)d\tau \bigg)\big(E_fE_h\big)^{-\frac{1}{2}} +\frac{1}{2}\mu_yE_f^{-\frac{1}{2}}E_h^{-\frac{3}{2}}\bigg(\partial_{h(\tau)}E_h\bigg)$

$\ \ \ \ \ \ \ \ \ = \bigg(\frac{1}{|dom(f)|}\int_{\infty}^{\infty} f(t-\tau)d\tau \bigg)\big(E_fE_h\big)^{-\frac{1}{2}} -\frac{1}{2}\mu_yE_f^{-\frac{1}{2}}E_h^{-\frac{3}{2}}\bigg(\partial_{h(\tau)}E_h\bigg)$

$\ \ \ \ \ \ \ \ \ = \big(E_fE_h\big)^{-\frac{1}{2}} \frac{1}{|dom(f)|}\int_{\infty}^{\infty} f(t-\tau)d\tau-\frac{1}{2}\mu_yE_f^{-\frac{1}{2}}E_h^{-\frac{3}{2}}\bigg(\partial_{h(\tau)}\int_{\infty}^{\infty} h^2(\tau)d\tau \bigg)$

$\ \ \ \ \ \ \ \ \ = \frac{1}{|dom(f)|}\big(E_fE_h\big)^{-\frac{1}{2}} \int_{\infty}^{\infty} f(t-\tau)d\tau-\mu_yE_f^{-\frac{1}{2}}E_h^{-\frac{3}{2}}\int_{\infty}^{\infty} h(\tau)d\tau$

$\ \ \ \ \ \ \ \ \ \propto \frac{1}{|dom(f)|}\int_{\infty}^{\infty} f(t-\tau)d\tau-\mu_yE_h^{-1}\int_{\infty}^{\infty} h(\tau)d\tau$

$\ \ \ \ \ \ \ \ \ = \int_{\infty}^{\infty} \bigg(\frac{1}{|dom(f)|}f(t-\tau)-h(\tau)\mu_yE_h^{-1}\bigg)d\tau$

At minimum $J$ we shouldn't assume $h(\tau) = 0$, we'll enforce $$\frac{1}{|dom(f)|}f(t-\tau)-h(\tau)\mu_yE_h^{-1} = 0$$

and pretty readily you get

$$h(\tau) = \frac{E_h}{\mu_y|dom(f)|}f(t-\tau)$$

Or most importantly

$$h(\tau) \propto f(t-\tau) \text{where} \ \ \tau \ \ \text{is time and} \ \ t \ \ \text{is the delay}$$

or, using perhaps better "variable names"

$$h(t) \propto f(t_d-t) \text{where} \ \ t \ \ \text{is time and} \ \ t_d \ \ \text{is the delay}$$

That is, if we want to maximize the correlation between our signal detector with impulse h(t), we better pick h(t) to be a time-reversed and time-shifted version of our signal of interest. In practice $t_d$ would probably be set to zero, as it just represents whenever the $f$ part that you're looking for finally arrives.

Under this chosen $h(t) \propto f(t_d - t)$, your original question makes more sense. The autocorrelation signal becomes proportional to the accumulated energy of $f(t)$ that is seen by your filter.

$y(t) = f(t) \ast h(t)$

$y(t) \propto \int f(\tau) h(t-\tau)d\tau$

$y(t) = \int f(\tau) f(t_d-(t+\tau))d\tau$

$y(t) = \int f(\tau) f(\tau + t_d - t)d\tau$

$y(t) = \int P_f(\tau + t_d - t)d\tau$